[英]MYSQL - PHP - Update value in database by using variable passed from submit button
我想知道是否有可能從數據庫中傳遞一個值(如下所示)到我的php文件中,並根據傳遞的參數更新相應的行值。 例如,當用戶單擊按鈕時,這將基於按鈕所在的ID更新數據庫內部的值。 謝謝。
<?php
// Selecting Database
include_once 'dbh.php';
//Here we fetch the data from the URL that was passed from our HTML form
$userEmail = $_POST['userEmail'];
$jobiD = $_POST['jobID'];
$sql = "UPDATE jobPost SET emailTeacher='$userEmail' WHERE jobID= ('".$jobID."');";
mysqli_query($conn, $sql);
?>
AJAX-
function myFunctionjobStatus() {
var jobID = document.getElementById("jobID").value;
//AJAX code to submit form.
$.ajax({
type: "POST",
url: "http://localhost:8888/EduSubOct/jobstatus.php",
data: { userEmail: localStorage.getItem("email"), 'jobID':
jobID },
cache: false,
success: function(html) {
alert("Request Sent");
}
});
}
PHP變量區分大小寫,您知道嗎?
$jobiD = $_POST['jobID'];
$作業ID應該是$作業ID,你用它在你的查詢字符串:
$sql = "UPDATE jobPost SET emailTeacher='$userEmail' WHERE jobID= ('".$jobID."');";
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